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Recently I became curious about the following question:

Let $V$ be a finite dimensional vector space over $k$ and let $A_1, \cdots, A_n: V \rightarrow V$ be a set of commuting maps. Question: describe the structure of $V$ as a module over $k[x_1, \cdots, x_n]$ where $x_i$ acts by $A_i$.

Since $V$ has a finite length as $k[x_1, \cdots, x_n]$-module, after we quotient by annihilator of $V$, we have a module over artinian ring (geometrically speaking, support is discrete set), so $V$ is isomorphic to direct sum of it's localizations at prime ideals (all primes are maximal in this case). Is there a description of each of the local components? We can assume that $k$ is algebraically closed, so that Nullstellensatz might help.

Thanks for your replies.

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As soon as $n>1$ we get a wild classification problem (in the sense of representation theory) and hence any kind of explicit description is hopeless. –  Torsten Ekedahl Feb 17 '11 at 18:26
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I don't understand this comment at all. –  David Hill Feb 17 '11 at 18:46
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You can generalize the argument of simultaneous diagonalization to reduce the problem. More precisely, you can split $V$ into a direct sum of invariant pieces, and on each piece each $A_i$ is the sum of a scalar and a nilpotent one. This reduces the problem to the case where all $A_i$ are nilpotent. –  Johannes Ebert Feb 17 '11 at 18:55
    
@Tortsten Ekedahl: can you please give a reference for the "wild classification problem" –  Mikhail Gudim Feb 17 '11 at 20:53
    
Johannes- The OP already said that; it's the same as his argument about reducing to localizations at prime ideal. –  Ben Webster Feb 18 '11 at 2:29
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4 Answers

up vote 1 down vote accepted

It is well known that representation theory of a (even commutative) Artinian $k$-algebra $R$ can be wild (meaning that one can embed $\mod(A)$ into $\mod(R)$ for any finite-dimensional, not necessarily commutative, $k$-algebra $A$). The easiest example is $k[x,y]/(x^2,xy^2,y^3)$, see Example 4 in this paper and the references there.

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I guess Torsten already said it in the comment! –  Hailong Dao Feb 18 '11 at 1:41
    
Thanks Hailong, now I see that this is hard problem. –  Mikhail Gudim Feb 18 '11 at 4:37
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Let me elaborate on Johannes Ebert's comment.

Since you are dealing with commuting operators, you can find a basis for $V$ so that the matrices for $A_1,\ldots,A_n$ are in Jordan normal form. So, $V$ decomposes as $V=V_1\oplus\cdots\oplus V_r$, where for each $i=1,\ldots,r$, there exists scalars $a_{ij}$, $j=1,\ldots,n$, such that $(x_j-a_{ij})^Nv=0$ for all $v\in V_i$ and $N\gg0$. Moreover, we may assume that each $V_i$ is indecomposable.

Now, we may assume $r=1$, and let $a_j$ be the generalized eigenvalues for the action of $x_j$ on $V$. By replacing $x_j$ by $x_j-a_j$, we may assume all $a_j$ are 0, and the $x_j$ are nilpotent.

Now, the problem is to identify commuting families of $n$ nilpotent operators acting on a finite dimensional vector space, say of dimension $N$. To do this, for each subset $S\subset\{ 1,\ldots,N-1\}$, consider the matrix $e_S=\sum_{i\in S}e_{i,i+1}$. Then, we are looking for a collection of subsets $S_1,\ldots,S_n$ such that $e_{S_i}e_{S_j}=e_{S_j}e_{S_i}$. If we consider the special case where $S_i$ and $S_j$ are intervals, we see that we must have either $S_i=S_j$ or $S_i\cup S_j$ consists of two disjoint intervals. To generalize, each $S_i=\bigcup S_{ik}$, where each $S_{ik}$ is an interval, but no $S_{ik}\cup S_{il}$ is an interval (or $S_i=\emptyset$).

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I think in my argument above reduction to the local case is the same as reducing to the case of nilpotent maps. (Module over local artinian ring is annihilated by some power of maximal ideal.) In case $n=1$ we can decompose each piece further because since we are looking at a module over the ring of the form $k[x]/(x-a)^n$ and this ring is self-injective we can split of cyclic submodules. This gives the usual Jordan decomposition. My question basically is: what if anything can be done when $n>1$?. Is there some technique from homological algebra that might be useful here? –  Mikhail Gudim Feb 17 '11 at 21:14
    
The argument above doesn't have a restriction on n. –  David Hill Feb 17 '11 at 22:01
    
I am confused. Do you claim that a family of n commuting nilpotent linear maps on a space of dimension N can be brought to the form you describe by a basis change? (This is definitely false.) If you do not make this claim, then you do provide some examples, but it does not advance the classification problem. Even when n=2, which is where most research was done, there are non-trivial statements. Some key words: pairs of commuting nilpotent matrices, (local) Hilbert scheme of a surface. –  t3suji Feb 17 '11 at 23:25
    
You are right. I'm not sure what this comment was about... –  David Hill Feb 18 '11 at 1:07
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I haven't read this paper, but it looks relevant: Principal nilpotent pairs in a semisimple Lie algebra I. To quote from the MathSciNet review:

The author notes that the general problem of classifying $G$-orbits through arbitrary nilpotent pairs is wild; even the set of orbits of maximal dimension has no structure of algebraic variety.

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These are not arbitrary nilpotent pairs. They commute. –  David Hill Feb 18 '11 at 1:07
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I thought the result that this is a wild problem was due to Ringel. However I have not been able to find the reference. Here is a reference for pairs of nilpotent elements (without the assumption that they commute).

MR2376281 (2009d:16016) Ringel, Claus Michael ; Schmidmeier, Markus . Invariant subspaces of nilpotent linear operators. I. J. Reine Angew. Math. 614 (2008), 1--52.

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